Evaluate : `int_(0)^(pi)(x)/(a^(2)cos^(2)x+b^(*2)sin^(2)x)dx`

Explanation

We have, `I =underset(0)overset(pi)int(x)/(a^(2)cos ^(2)x+b^(2)sin^(2)x)dx" "...(1)`
`I =underset(0)overset(pi)int(pi-x)/(a^(2)cos^(2)(pi-x)+b^(2)sin^(2)(pi-x))dx`
`(because underset(o)overset(a)intf(x)dt=underset(o)overset(a)intf(a-x)dx)`
`I=underset(o)overset(pi)int(pi-x)/(a^(2)cos^(2)x+b^(2)sin^(2)x)dx" "...(2)`
On adding equations (1) a nd (2), we get
`21 =underset(0)overset(pi)int(x+pi-x)/(a^(2)cos^(2)x+b^(2)sin^(2)x)dx`
Using the property, `underset(0)overset(pi)intf(x)*dx=underset(0)overset(pi)int[f(x)+(2a-x)]dx`
`therefore2I=pi[underset(0)overset(pi)int(1)/(a^(2)cos^(2)x+b^(2)sin^(2)x)dx+underset(0)overset(pi//2)int(1)/(a^(2)cos^(2)(pi-x)+b^(2)sin^(2)(pi-x))dx]`
`I=pi underset(0)overset(pi//2)int(dx)/(a^(2)cos^(2)x(1+(b^(2))/(a^(2))tan^(2)x))`
`=(pi)/(a^(2))underset(0)overset(pi//2)int(sec^(2)x dx)/(1+(b^(2))/(a^(2))tan^(2)x)`
[By putting `u=b/atanx, du =b/asec^(2)x dx` when `x=0, u=0`
when `x=pi/2, u=oo`
`=(pi)/(a^(2))underset(0)overset(oo)int((a)/(b)du)/(a+u^(2))`
`=(pi)/(ab)underset(0)overset(oo)int(du)/(1+u^(2))=(pi)/(ab)[tan^(-1)(u)]_(0)^(oo)+c`
`=(pi)/(ab)[tan^(-1),ootan^(-1)0]+c`
`=(pi)/(ab)[(pi)/(2)-0]+c`
`=(pi^(2))/(2ab)`